Problem: You have found the following ages (in years) of all 6 snakes at your local zoo: $ 31,\enspace 22,\enspace 34,\enspace 10,\enspace 2,\enspace 25$ What is the average age of the snakes at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 snakes at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{31 + 22 + 34 + 10 + 2 + 25}{{6}} = {20.7\text{ years old}} $ Find the squared deviations from the mean for each snake. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $31$ years $10.3$ years $106.09$ years $^2$ $22$ years $1.3$ years $1.69$ years $^2$ $34$ years $13.3$ years $176.89$ years $^2$ $10$ years $-10.7$ years $114.49$ years $^2$ $2$ years $-18.7$ years $349.69$ years $^2$ $25$ years $4.3$ years $18.49$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{106.09} + {1.69} + {176.89} + {114.49} + {349.69} + {18.49}} {{6}} $ $ {\sigma^2} = \dfrac{{767.34}}{{6}} = {127.89\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{127.89\text{ years}^2}} = {11.3\text{ years}} $ The average snake at the zoo is 20.7 years old. There is a standard deviation of 11.3 years.